• Physics 9702 Doubts | Help Page 63

Question 353: [Radioactive decay]
Because of the random nature of the radioactive emission process, the count-rate recorded by a Geiger-Muller tube or solid-state detector from a supposedly constant source is subject to statistical fluctuations. For example, if the total count recorded from such a source in a given time is N, the variability of repeated measurements is of the order of √N. For how many counts should counting continue in order to obtain the mean count rate to a precision of about 1%?
A 100              B 1 000           C 10 000                     D 100 000                   E 1 000 000

Reference: Past Exam Paper – N83 / II / 35



Solution 353:
Answer: C.
The variability of repeated measurements (ΔN) is of the order of √N.
(√N / N) x 100% = 1%
√N / N = 0.01
N^0.5 N^-1 = N^-0.5 = 0.01
1 / √N = 0.01
N = (1 / 0.01)² = 100² = 10 000
 








Question 354: [Nuclear Physics]
A stationary ²³⁸U nucleus decays by α emission generating a total kinetic energy T.

²³⁸₉₂U   - - - >               ²³⁴₉₀Th +          ⁴₂α

What is the kinetic energy of the α particle?
A slightly less than T / 2
B T / 2
C slightly less than T
D T
E slightly greater than T

Reference: Past Exam Paper – N 90 / I / 6



Solution 354:
Answer: C.
The ²³⁸U nucleus is initially stationary, so it has zero momentum.

From the law of conservation of momentum, the sum of momentum before the decay should be equal to the sum of momentum after the decay.

After the decay,
Let momentum of ²³⁴₉₀Th = Mv_Th = (234u) v_Th 
and let the momentum of ⁴₂α = mv_α = (4u) v_α

From the law of conservation of momentum,
(234u) v_Th + (4u) v_α = 0

Considering the magnitudes,
(234u) v_Th + (4u) v_α
Speed of α-particle, v_α = (234u) v_Th / (4u)

So, the speed of alpha-particle is much greater than that of Th. Since kinetic energy depends on the square of speed, the kinetic energy of the α particle is then just slightly less than T.










Question 355: [Simple harmonic motion]
A particle undergoes simple harmonic motion on a straight line with amplitude 5.0 cm and frequency 2.0 Hz. At time t = 0, it starts to move to the right from the equilibrium position. Determine
(a) the displacement at t = 0.10 s and t = 0.40 s;
(b) the instants when the displacements of the object are 1.0 cm and – 2.0 cm for the first time.



Solution 355:
The equation describing the displacement x of a particle undergoing simple harmonic motion with respect to time t is given by

x = x₀ sin (ωt)

where x₀ is the amplitude (maximum displacement) (= 5.0cm) and ω is the angular frequency of the simple harmonic motion

Angular frequency ω = 2π /T = 2πf = 2π (2) = 4π

So, the equation can be simplified as
x = 5 sin (4πt)

Note that the argument of the sine function should be in radians.

(a)
x = 5 sin (4πt)

When time t = 0.10s,
Displacement x = 5 sin (4π (0.10)) = 4.76cm

When time t = 0.40s,
Displacement x = 5 sin (4π (0.40)) = – 4.76cm

(b)
x = 5 sin (4πt)

To find the time, we need to make t the subject of formula.
sin (4πt) = x / 5
4πt = sin^-1 (x/5)
Time t = [sin^-1 (x/5)] / 4π
When displacement x = 1.0cm
Time t = [sin^-1 (1.0/5)] / 4π = 0.016s = 0.02s

When displacement x = – 2.0cm
Time t = [sin^-1 (– 2.0/5)] / 4π = 0.016s = 0.28s









Question 356: [Electricity > Electric field strength]

Two large flat metal plates A and B are placed 9.0 cm apart in a vacuum, as illustrated.

A potential difference of 450 V is maintained between plates by means of a battery.

(a)

(i) On Fig, draw arrow to indicate the direction of the electric field between plates A and B.

(ii) Calculate electric field strength between A and B.

(b) An electron is released from rest at surface of plate A.

(i) Show that change in electric potential energy in moving from plate A to plate B is 7.2 ×10^–17J.

(ii) Determine speed of the electron on reaching plate B.

(c) On axes of Fig, sketch graph to show the variation with distance d from plate A of the speed v of the electron.

Reference: Past Exam Paper – November 2003 Paper 2 Q5



Solution 356:

(a)

(i) The arrow is from B towards A.

(ii) Electric field strength, E = V / d = 450 / (90x10^-2) = 5.0x10³ NC^-1

(b)

(i)

Change in electric potential energy = qV or Eqd

Change in electric potential energy = (1.6x10^-19) (450) = 7.2x10^-17 J

(ii)

{The change in electric potential energy is converted into kinetic energy of the electron.}

Kinetic energy, E_k = ½ mv²

7.2x10^-17 = 0.5 (9.1x10^-31) v²

Speed v = 1.26x10⁷ ms^-1

(c)

The line starts from the origin, and is curved in the correct direction, but not ‘level out’